Page 497 - Electromagnetics

We produce a lack of z-dependence in ψ by letting k z = 0 and choosing a z = 0. Moreover,
2
2
k =−k since Laplace’s equation requires k = 0. This leads to three possibilities. If
x y
k x = k y = 0, we have the product solution
ψ(x, y) = (a x x + b x )(a y y + b y ). (A.106)
If k y is real and nonzero, then

ψ(x, y) = (A x e −k y x + B x e k y x )(A y e jk y y + B y e − jk y y ). (A.107)
Using the relations
u −u u −u
sinh u = (e − e )/2 and cosh u = (e + e )/2 (A.108)
along with (A.101), we can rewrite (A.107) as

ψ(x, y) = (A x sinh k y x + B x cosh k y x)(A y sin k y y + B y cos k y y). (A.109)
(We can reuse the constant names A x , B x , A y , B y , since the constants are unknown at
this point.) If k x is real and nonzero we have

ψ(x, y) = (A x sin k x x + B x cos k x x)(A y sinh k x y + B y cosh k x y). (A.110)
Consider the problem consisting of Laplace’s equation
2
∇ V (x, y) = 0 (A.111)
holding in the region 0 < x < L 1 , 0 < y < L 2 , −∞ < z < ∞, together with the boundary
conditions
V (0, y) = V 1 , V (L 1 , y) = V 2 , V (x, 0) = V 3 , V (x, L 2 ) = V 4 .

The solution V (x, y) represents the potential within a conducting tube with each wall
held at a different potential. Superposition applies: since Laplace’s equation is linear
we can write the solution as the sum of solutions to four different sub-problems. Each
sub-problem has homogeneous boundary conditions on one independent variable and
inhomogeneous conditions on the other, giving a Sturm–Liouville problem in one of the
variables. For instance, let us examine the solutions found above in relation to the sub-
problem consisting of Laplace’s equation (A.111) in the region 0 < x < L 1 , 0 < y < L 2 ,
−∞ < z < ∞, subject to the conditions

V (0, y) = V (L 1 , y) = V (x, 0) = 0, V (x, L 2 ) = V 4 = 0.
First we try (A.106). The boundary condition at x = 0 gives

V (0, y) = (a x (0) + b x )(a y y + b y ) = 0,

which holds for all y ∈ (0, L 2 ) only if b x = 0. The condition at x = L 1 ,
V (L 1 , y) = a x L 1 (a y y + b y ) = 0,

then requires a x = 0. But a x = b x = 0 gives V (x, y) = 0, and the condition at y = L 2
cannot be satisfied; clearly (A.106) was inappropriate. Next we examine (A.109). The
condition at x = 0 gives
V (0, y) = (A x sinh 0 + B x cosh 0)(A y sin k y y + B y cos k y y) = 0,

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